Putnam 2003
A1. Let n be a fixed positive integer. How many ways are there to write n as a sum of
positive integers,
n = a1 + a2 + . . . + ak ;
with k an arbitrary positive integer and a1 ≤ a2 ≤ · · · ≤ a1 + 1? For example, with n = 4,
there are four ways: 4, 2 + 2, 1 + 1 + 2, 1 + 1 + 1 + 1.
Solution. We use induction to prove that the number N (n) of ways is n. For n = 1 this is
clear. By gathering the equal terms (either a1 or a1 + 1), each equation n = a1 + a2 + . . . + ak
for given n and k can be uniquely rewritten in the form
E (n, k, a, r) : n = ra + (k − r) (a + 1)
for some r ∈ {1, . . . , k} which depends uniquely on k, a,and n (indeed, r = k (a + 1) − n).
Let
©
ª
En := E (n, k, a, r) : E (n, k, a, r) holds for some (k, a, r) ∈ Z3+ .
Note that
n = ra + (k − r) (a + 1) ⇒
n + 1 = (r − 1) a + (k − (r − 1)) (a + 1) .
Thus, we have a map
F : En → En+1 ,
defined by
F (E (n, k, a, r)) :=
½
E (n + 1, k, a + 1, k) r = 1
E (n + 1, k, a, r − 1) 2 ≤ r ≤ k ≤ n.
Note that F is clearly 1 − 1. It is not onto, since it misses the value
E (n + 1, n + 1, 1, n + 1) ,
but this is the only value it misses, since
½
F (E (n, k, a, s + 1)) if s ≤ k − 1
E (n + 1, k, a, s) =
F (E (n, k, a − 1, k)) if s = k ≤ n.
Hence, N (n) = n ⇒ N (n + 1) = n + 1.
A2. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be nonnegative real numbers. Show that
1
1
1
(a1 a2 · · · an ) n + (b1 b2 · · · bn ) n ≤ ((a1 + b1 ) (a2 + b2 ) · · · (an + bn )) n .
1
Solution. Using the Lagrange multiplier method, for yi ≥ 0 (i = 1, . . . n), the maximum
of the function y1 y2 · · · yn with the constraint y1 + y2 + · · · + yn = 1, is achieved when
y1 = y2 = · · · = yn = n1 . Thus,
1
(y1 y2 · · · yn ) n ≤
1
if y1 + y2 + · · · + yn = 1 and yi ≥ 0 (i = 1, . . . n).
n
For xi ≥ 0 (i = 1, . . . n) and s = x1 + x2 + · · · + xn , taking yi = xi /s, we obtain
³x x
xn ´ n1
1
1 2
···
or
≤
s s
s
n
1
1
(x1 x2 · · · xn ) n ≤
(x1 + x2 + · · · + xn ) ,
n
(1)
which says that the geometric mean is not greater than the arithmetic mean. We may assume
that ai + bi > 0 for all i, since the result is clearly true if ai + bi = 0 for some i. Apply (1)
i
for xi = aia+b
to get
i
µ
a2
an
a1
·
···
a1 + b1 a2 + b2
an + bn
Now apply it for xi =
µ
bi
ai +bi
¶ n1
1
≤
n
µ
a2
an
a1
+
+ ··· +
a1 + b1 a2 + b2
an + bn
¶
.
¶ n1
1
≤
n
µ
b2
bn
b1
+
+ ··· +
a1 + b1 a2 + b2
an + bn
¶
.
to get
b2
bn
b1
·
···
a1 + b1 a2 + b2
an + bn
Adding the above, we get
µ
a1
a2
an
·
···
a1 + b1 a2 + b2
an + bn
¶ n1
+
µ
b1
b2
bn
·
···
a1 + b1 a2 + b2
an + bn
¶ n1
≤
1
Now multiply by ((a1 + b1 ) (a2 + b2 ) · · · (an + bn )) n .
A3. Find the minimum value of
|sin x + cos x + tan x + cot x + sec x + csc x|
for real numbers x.
Solution.
1
sin x cos x sin2 x + cos2 x
+
=
=
tan x + cot x =
cos x sin x
cos x sin x
cos x sin x
1
1
sin x + cos x
sec x + csc x =
+
=
.
cos x sin x
cos x sin x
2
1
·n=1
n
√
√
Let u = sin x + cos x and note that − 2 ≤ u ≤ 2. Then
u2 = sin2 x + 2 cos x sin x + cos2 x = 1 + 2 cos x sin x
u2 − 1
.
⇒ cos x sin x =
2
Thus,
sin x + cos x + tan x + cot x + sec x + csc x = u +
= u+
f 0 (u) = 1 −
√
2
=
0
for
u
=
1
±
2
(u − 1)2
³
√ ´
√
f 1± 2 =1± 2+
We check the endpoints:
We have
2
2u
+ 2
−1 u −1
2 (u + 1)
2
2 (u + 1)
=
u
+
=
u
+
:= f (u) .
u2 − 1
(u + 1) (u − 1)
u−1
We have
Now,
u2
√
2
√
= 1 ± 2 2.
1± 2−1
¢
¡√
³√ ´
√
√
2 2+1
2
¢ ¡√
¢
2 =
2+ √
f
= 2 + ¡√
2−1
2−1
2+1
³√
´
√
√
=
2+2
2 + 1 = 2 + 3 2 and
¢
¡√
³ √ ´
√
√
2 2−1
2
¢ ¡√
¢
f − 2 = − 2+ √
= − 2 − ¡√
− 2−1
2+1
2−1
³√
´
√
√
= − 2−2
2 − 1 = 2 − 3 2.
n¯
√ ¯¯ ¯¯
√ ¯¯ ¯¯
√ ¯¯o
√ ¯¯ ¯¯
¯
min ¯1 − 2 2¯ , ¯1 + 2 2¯ , ¯2 − 3 2¯ , ¯2 + 3 2¯
n √
³ √
´ √
o
√
√
= min 2 2 − 1, 3 2 − 2 = 2 2 − 1 + 2 − 1 = 2 2 − 1 ≈ 1.828.
√
Thus, the answer is 2 2 − 1.
A4. Suppose that a, b, c, A, B, C are real numbers, a 6= 0 and A 6= 0, such that
¯ ¯
¯
¯ 2
¯ax + bx + c¯ ≤ ¯Ax2 + Bx + C ¯
for all real numbers x. Show that
¯ 2
¯ ¯
¯
¯b − 4ac¯ ≤ ¯B 2 − 4AC ¯ .
3
Solution. By replacing a, b, c by −a, −b, −c, we may assume that a > 0, and similarly we
may assume that A > 0. Note that none of the above absolute values are affected by such
replacements. For large x
¯
¯
¯
¯
¯ ¯
¯
¯ 2
¯ax + bx + c¯ ≤ ¯Ax2 + Bx + C ¯ ⇒ ¯a + bx−1 + cx−2 ¯ ≤ ¯A + Bx−1 + Cx−2 ¯
⇒ a ≤ A, taking the limit as x → ∞.
Let D := B 2 − 4AC and d := b2 − 4ac.
There are three cases
(i) D ≥ 0, (ii) D < 0 and d > 0, (iii) D < 0 and d < 0
Case (i) D ≥ 0: If D ≥ √0, then Ax2 + Bx + C has real zeros r2 ≥ r1 , and by the
quadratic formula, r2 − r1 = AD . Note that r1 and r2 are also zeros of ax2 + bx + c, since
|ax2 + bx + c| ≤ |Ax2 + Bx + C|. Thus,
√
√
√
√
d
D
a D √
= r2 − r1 =
⇒ d=
≤ D ⇒ d ≤ D.
a
A
A
Case (ii) D < 0 and d > 0. If D < 0, then 0 < Ax2 + Bx + C and
¯ ¯
¯
¯ 2
¯ax + bx + c¯ ≤ ¯Ax2 + Bx + C ¯ = Ax2 + Bx + C
Thus,± (ax2 + bx + c) ≤ Ax2 + Bx + C and so
(A ± a) x2 + (B ± b) x + (C ± c) ≥ 0
Hence
(B ± b)2 − 4 (A ± a) (C ± c) ≤ 0,
¢ ¡
¢
¡ 2
B − 4AC + b2 − 4ac ≤ ± (−2Bb + 4Ac + 4aC) .
Since one of ± (−2Bb + 4Ac + 4aC) is ≤ 0, we have
D + d ≤ 0 and so |d| = d ≤ |D| .
Case (iii) D < 0 and d < 0: In this case, 0 ≤ ax2 + bx + c ≤ Ax2 + Bx + C and
consequently
©
ª
©
ª
min ax2 + bx + c ≤ min Ax2 + Bx + C
Now,
µ ¶2
µ ¶
© 2
ª
4ac − b2
−d
−b
−b
min ax + bx + c = a
+c=
=
.
+b
2a
2a
4a
4a
4
Thus,
|d|
−d
−D
|D|
a
=
≤
=
and |d| ≤ |D| ≤ |D| .
4a
4a
4A
4A
A
A5. A Dyck n-path is a lattice path of n upsteps (1, 1) and n downsteps (1, −1) that starts
at the origin O and never dips below the x-axis. A return is a maximal sequence of contiguous
downsteps that terminates on the x-axis. For example, the Dyck 5-path illustrated has two
returns, of length 3 and 1 respectively.
Show that there is a one-to-one correspondence between the Dyck n-paths with no return
of even length and the Dyck (n − 1)-paths.
Solution. Let us denote a Dyck n-path by its sequence of steps v1 , v2 , . . . , v2n , where
vi = (1, 1) or (1, −1). Denote the set of Dyck n-paths by Dn . We define a function
F : Dn−1 → Dn (∼ e) := Dn \ {paths in Dn with a return of even length}
If v1 , v2 , . . . , v2(n−1) ∈ Dn−1 (∼ e), then we set
¡
¢
F v1 , v2 , . . . , v2(n−1) = (1, 1), (1, −1), v1 , v2 , . . . , v2(n−1) ∈ Dn (∼ e) .
If v1 , v2 , . . . , v2(n−1) ∈ Dn−1 \ Dn−1 (∼ e) (i.e., a Dyck (n − 1)-path with a return of even
length), let the final return of even length end with the step vf . Note that in the Dyck
n-path (1, 1), v1 , v2 , . . . , vf , (1, −1), vf +1 , . . . , v2(n−1) the original subpath v1 , v2 , . . . , vf has
been translated up one unit and to the right one unit so that (1, 1), v1 , v2 , . . . , vf , (1, −1) has
only one return, namely a final return of odd length, while (by choice of vf ) vf +1 , . . . , v2(n−1)
has no return of even length. Thus, for v1 , v2 , . . . , v2(n−1) ∈ Dn−1 \ Dn−1 (∼ e), we set
¢
¡
F v1 , v2 , . . . , v2(n−1) := (1, 1), v1 , v2 , . . . , vf , (1, −1), vf +1 , . . . , v2(n−1) ∈ Dn (∼ e) .
To show that F is a bijection, we need to find an inverse, say G, for F . Let v1 , v2 , . . . , v2n ∈
Dn (∼ e) and let v1 , v2 , . . . , v2i be the first Dyck subpath of v1 , v2 , . . . , v2n . Then set
G (v1 , v2 , . . . , v2n ) = v2 , . . . , v2i−1 , v2i+1 , . . . , v2n .
Here, if i = 1, v2 , . . . , v2i−1 is empty, and G (v1 , v2 , . . . , v2n ) = v3 , . . . , v2n . Note that since
v1 , v2 , . . . , v2n ∈ Dn (∼ e), the first return (if any) of G (v1 , v2 , . . . , v2n ) = v2 , . . . , v2i−1 , v2i+1 , . . . , v2n ,
ends in v2i−1 , has even length, and is the last return of even length, in which case F (G (v1 , v2 , . . . , v2n )) =
5
v1 , v2 , . . . , v2n . If v2 , . . . , v2i−1 , v2i+1 , . . . , v2n has no return of even length, then it must have
been that i = 2, in ¡which
¡ case we also have
¢¢ F (G (v1 , v2 , . . . , v2n )) = (1, 1), (1, −1), v3 , . . . , v2n =
v1 , . . . , v2n . For G F v1 , v2 , . . . , v2(n−1) , note that if v1 , v2 , . . . , v2(n−1) ∈ Dn−1 (∼ e), then
¢¢
¡
¢
¡ ¡
G F v1 , v2 , . . . , v2(n−1) = G (1, 1), (1, −1), v1 , v2 , . . . , v2(n−1) = v1 , v2 , . . . , v2(n−1) .
If v1 , v2 , . . . , v2(n−1) ∈ Dn−1 \ Dn−1 (∼ e), then
¢¢
¡
¢
¡ ¡
= G (1, 1), v1 , v2 , . . . , vf , (1, −1), vf +1 , . . . , v2(n−1)
G F v1 , v2 , . . . , v2(n−1)
= v1 , v2 , . . . , v2(n−1) ,
as required.
A6. For a set S of nonnegative integers, let rS (n) denote the number of ordered pairs
(s1 , s2 ) such that s1 ∈ S, s2 ∈ S, s1 6= s2 , and s1 + s2 = n. Is it possible to partition the
nonnegative integers into two sets A and B in such a way that rA (n) = rB (n) for all n?
Solution. Let 0 ∈ A = {0, . . .}. Then
1
2
3
4
5
6
7
=
=
=
=
=
=
=
0 + 1 ⇒ 1 ∈ B = {1, . . .} ,
0 + 2 ⇒ 2 ∈ B = {1, 2, . . .} ,
0 + 3 = 1 + 2 ⇒ 3 ∈ A = {0, 3, . . .} ,
0 + 4 = 1 + 3 ⇒ 4 ∈ B = {1, 2, 4, . . .} ,
0 + 5 = 1 + 4 = 2 + 3 ⇒ 5 ∈ A = {0, 3, 5, . . .} ,
0 + 6 = 1 + 5 = 2 + 4 ⇒ 6 ∈ A = {0, 3, 5, 6, . . .} ,
0 + 7 = 1 + 6 = 2 + 5 = 3 + 4 ⇒ 7 ∈ B = {1, 2, 4, 7, . . .} .
Note that thus far A consists of the whole numbers (Z+ ) with an even number of ones in
their base 2 representation, whereas B consists of the whole numbers with an odd number
of ones in their base 2 representation. Let’s prove the conjecture. Let n ≥ 0 and suppose
n = a1 + a2 where a1 , a2 ∈ A, a1 6= a2 .
Now the binary reps of a1 and a2 differ in some first digit, say from the right. Change that
digit in each of a1 and a2 , to obtain b1 and b2 ∈ B. Note that b1 + b2 = a1 + a2 = n. Thus,
we have a bijection
{(a1 , a2 ) : a1 + a2 = n with a1 , a2 ∈ A, a1 6= a2 }
←→ {(b1 , b2 ) : b1 + b2 = n with b1 , b2 ∈ B, b1 6= b2 } ,
showing that rA (n) = rB (n) for all n ∈ Z+ .
B1. Do there exist polynomials a(x), b(x), c(y), d(y) such that
1 + xy + x2 y 2 = a(x)c(y) + b(x)d(y)
6
holds identically?
Solution. No. Choosing y = 0, y = ±1, we get
1 = c(0)A(x) + d(0)B(x)
1 + x + x2 = c(1)A(x) + d(1)B(x)
1 − x + x2 = c(−1)A(x) + d(−1)A(x),
where A(x) and B(x) are the truncations of a(x) and b (x) to polynomials of degree less than
3; note that any higher degree terms in a(x) and b (x) must cancel on the right sides. Since
{1, 1 + x + x2 , 1 − x + x2 } is a basis of the vector space P2 of polynomials in x of degree
less than 3 and each of 1, 1 + x + x2 , 1 − x + x2 is a linear combination of A(x) and B(x),
{A(x), B(x)} spans P2 , but dim P2 = 3, and so a spanning set of P2 must have at least 3
elements.
B2. Let n be a positive integer. Starting with the sequence 1, 12 , 13 , . . . n1 , form a new
5
2n−1
sequence of n − 1 entries, 34 , 12
, . . . , 2n(n−1)
, by taking the averages of two consecutive entries
in the first sequence. Repeat the averaging of neighbors on the second sequence to obtain a
third sequence of n − 2 entries and continue until the final sequence produced consists of a
single number xn . Show that xn < n2 .
Solution. For an infinite sequence a0 , a1 , . . . , the sequence of first averages is
1
2
(a0 + a1 ) , 12 (a1 + a2 ) , 12 (a2 + a3 ) , . . . ,
the sequence of second averages is
¢ 1 ¡1
¢
¡
1 1
1
1
(a
+
a
)
+
(a
+
a
)
,
(a
+
a
)
,
(a
+
a
)
,...
0
1
1
2
1
2
2
3
2 2
2
2 2
2
1
1
= 22 (a0 + 2a1 + a2 ) , 22 (a1 + 2a2 + a3 ) , . . . ,
the sequence of third averages is
=
1
23
1
23
((a0 + 2a1 + a2 ) + (a1 + 2a2 + a3 )) , . . .
(a0 + 3a1 + 3a2 + a3 ) , . . .
By induction, the first term of the sequence of k-th averages is
à k
!
1 X
k!
ai .
2k i=0 i! (k − i)!
The value that we seek to estimate, namely the first term of the sequence of (n − 1)-th
averages of 1, 12 , 13 , . . . n1 , 0, 0, . . . , is then
à n−1
à n−1
!
!
X (n − 1)!
X
1
1
n!
1
=
2n−1 i=0 i! (n − 1 − i)! 1 + i
n2n−1 i=0 (i + 1)! (n − (i + 1))!
2
1
n
.
(2
−
1)
≤
n2n−1
n
=
7
B3. Show that for each positive integer n,
n! =
n
Y
i=1
¡
¥ ¦¢
lcm 1, 2, . . . , ni
(Here lcm denotes the least common multiple, and bxc denotes the greatest integer ≤ x.)
Solution. We use induction, first noting that the case n = 1 holds. We need to show that
¡
¥ n ¦¢
Qn
i=1 lcm 1, 2, . . . , i
¡
¥ n−1 ¦¢.
n = Qn−1
lcm
1,
2,
.
.
.
,
i=1
i
¡
¥ ¦¢
Q
Q
Since lcm 1, 2, . . . , ni = 1 when i = n, we may replace ni=1 by n−1
i=1 in the numerator,
whence we are to show that
Ã
¡
¥ ¦¢ !
n−1
Y
lcm 1, 2, . . . , ni
¡
¥
¦¢
(2)
n=
lcm 1, 2, . . . , n−1
i
i=1
¥ n ¦ ¥ n−1 ¦
¥ n ¦ ¥ n−1 ¦
¥ n ¦ ¥ n−1 ¦
Note
that
either
=
or
=
+
1.
If
= i , the i-th factor of 2 is 1.
i
i
i
i
i
¥ n ¦ ¥ n−1 ¦
n
If i = i + 1, then it must be that i is an integer. We will use the fact that the lcm
of a set of positive integers is the product of their maximal prime power factors. If ni is not
the power
prime
power factors of the numbers
¥ nof
¦ a prime, then all of the exponents of ¥the
¦
n−1
1, 2, . . . , i are no greater than those of 1, 2, . . . , i in which case the i-th factor of 2
is 1. If ni is pm for
some
p, then the i-th factor of 2 is p, since pm−1 is among the
¥ n−1
¦ prime
numbers 1, 2, . . . , i , but pm (and its multiples) is not. Thus, the non-unit factors of the
right side of 2 are just the prime factors of n each repeated according to multiplicity. Hence
the right side of 2 is the prime factorization of n which of course equals n.
B4. Let f (z) = az 4 + bz 3 + cz 2 + dz + e = a (z − r1 ) (z − r2 ) (z − r3 ) (z − r4 ) where
a, b, c, d, e are integers, a 6= 0. Show that if r1 +r2 is a rational number, and if r1 +r2 6= r3 +r4 ,
then r1 r2 is a rational number.
Solution. Note that
(z − r1 ) (z − r2 ) (z − r3 ) (z − r4 )
= z 4 − (r1 + r2 + r4 + r3 ) z 3 + (r1 r2 + r3 r4 + r2 r4 + r2 r3 + r1 r4 + r1 r3 ) z 2
− (r2 r3 r4 + r1 r3 r4 + r1 r2 r4 + r1 r2 r3 ) z + r1 r2 r3 r4
Thus,
(r1 + r2 ) + (r3 + r4 )
(r1 + r2 ) (r3 + r4 ) + r1 r2 + r3 r4
r3 r4 (r2 + r1 ) + r1 r2 (r4 + r3 )
(r1 r2 ) (r3 r4 )
=
=
=
=
r1 + r2 + r4 + r3 = −b/a
r1 r2 + r3 r4 + r2 r4 + r2 r3 + r1 r4 + r1 r3 = c/a
r2 r3 r4 + r1 r3 r4 + r1 r2 r4 + r1 r2 r3 = −d/a
r1 r2 r3 r4 = e/a
8
and the left sides are then rational. With s = r1 + r2 , t = r3 + r4 , u = r1 r2 and v = r3 r4 ,
these become
s+t
st + u + v
vs + ut
uv
=
=
=
=
−b/a
c/a
−d/a
e/a.
We conclude that t is rational, and u + v = c/a − st is rational. Then s (u + v) is rational
and
u (t − s) = vs + ut − s (u + v) = −d/a − s (u + v)
is then rational. If t 6= s, then
r1 r2 = u =
−d/a − s (u + v)
t−s
is rational as required.
B5. Let A, B and C be equidistant points on the circumference of a circle of unit radius
centered at O, and let P be any point in the circle’s interior. Let a, b, c be the distances
from P to A, B, C respectively. Show that there is a triangle with side lengths a, b, c, and
that the area of this triangle depends only on the distance from P to O.
Solution. Choose coordinates in the complex plane so that A = 1, B = e2πi/3 , C = e4πi/3 .
If β = e2πi/3 , then C = β 2 and A = β 3 = 1. Let P = z. Then
¯
¯
a = |z − 1| , b = |z − β| , and c = ¯z − β 2 ¯ .
To show that a, b, c are side lengths of a triangle we need unit complex numbers α1 , α2 , α3
so that
¡
¢
α1 (z − 1) + α2 (z − β) + α3 z − β 2 = 0, or equivalently,
¡
¢
(α1 + α2 + α3 ) z − α1 + α2 β + α3 β 2 = 0
2
¡We know
¢ that 1 + β + β = 0. ¡For (α12,¢α2 , α3 ), it is then natural to try permutations of
2
1, β, β . Indeed, (α1 , α2 , α3 ) = 1, β, β (or any other permutation) works fine, since
α1 + α2 β + α3 β 2 = 1 + β 2 + β 4 = 1 + β 2 + β = 0.
For a triangle with vector sides (a, b) and (c, d), the area ∆ is given by
¯
¶¯
µ
¯
¯
a
b
¯ = |bc − ad| = |Im ((a + ib) (c − id))|
2 · ∆ = ¯¯det
c d ¯
¯ ³
¯³
´¯
´¯
¯
¯ 1¯
¯
= ¯Im (a + ib) (c + id) ¯ = 2 ¯ (a + ib) (c + id) − (a + ib) (c + id) ¯
9
The area of the triangle above with sides z − 1, β (z − β) (and β 2 (z − β)) is then (noting
that β̄ = β 2 )
¯³
´¯
¯
1¯
(z
−
1)
β
(z
−
β)
−
(z
−
1)β
(z
−
β)
¯
¯
4
¯
¡
¢
¢¯
¡
= 1 ¯ (z − 1) β 2 z̄ − β 2 − (z̄ − 1) β (z − β) ¯
4
1
4
1
4
1
4
1
4
=
=
=
=
since
|((z − 1) (β z̄ − 1) − (z̄ − 1) (z − β))| |β|
|((βzz̄ − β z̄ − z + 1) − (z z̄ − z − β z̄ + β))|
¯¡¡
¢ ¡
¢¢¯
¯ β |z|2 + 1 − |z|2 + β ¯
¯¡
¢¢¯
¡
¡
¢
¯ (β − 1) |z|2 − 1 ¯ = 1 |(β − 1)| 1 − |z|2 =
4
√
3
4
¡
¢
1 − |z|2 ,
|(β − 1)|2 = (cos (2π/3) − 1)2 + sin2 (2π/3) = 2 − 2 cos (2π/3) = 3.
B6. Let f (x) be a continuous real-valued function defined on the interval [0, 1]. Show
that
Z 1
Z 1Z 1
|f (x) + f (y)| dxdy ≥
|f (x)| dx.
0
0
0
+
−
Solution.
Let A = {x ∈R[0, 1] : f (x) > 0} and let A = {x ∈ [0, 1] : f (x) ≤ 0} . For I + :=
R
f (x) dx and I − := − A− f (x) dx, we have
A+
Z
Z
Z 1
|f (x)| dx =
f (x) dx −
f (x) dx = I + + I − .
A−
A+
0
Also,
Z Z
=
+ ×A+
Z ZA
We have
¡ ¢
f (x) + f (y) dxdy = m A+
+
+
¡A +×A
¢ +
= 2m A
and similarly
|f (x) + f (y)| dxdy
I
Z Z
A− ×A−
Z Z
A+ ×A−
≥ ±
µZ Z
Z
A+
¡ ¢
f (x) dx + m A+
Z
¡ ¢
|f (x) + f (y)| dxdy = 2m A− I − .
|f (x) + f (y)| dxdy
¶
|f (x)| − |f (y)| dxdy
¶
µZ Z
Z Z
f (x) dxdy −
|f (y)| dxdy
= ±
A+ ×A−
A+ ×A−
¡ ¢ ¢
¡ ¡ ¢
= ± m A− I + − m A+ I −
A+ ×A−
10
A+
f (y) dy
Similarly,
Z Z
|f (x) + f (y)| dxdy
¡
¡ ¢ ¢
≥ ± m A I − m A− I +
A− ×A+
¡ +¢ −
Thus,
Z 1Z
Z0
Z
0
1
|f (x) + f (y)| dxdy
|f (x) + f (y)| dxdy
¡ ¢ ¢ ¡ ¡ ¢
¡ ¢ ¢
≥ 2m A I + 2m A I ± m A I − m A+ I − ±0 m A+ I − − m A− I +
¡ ¢
¡ ¢
¡ ¢
¡ ¢
¡ ¢
¡ ¢
= 2m A+ I + + 2m A− I − ± m A− I + ∓ m A+ I − ±0 m A+ I − ∓0 m A− I +
¡ ¡ ¢
¡ ¢¢
¡ ¡ ¢
¡ ¢¢
¡ ¢
¡ ¢
= 2m A+ ± m A− ∓0 m A− I + + 2m A− ∓ m A+ ±0 m A+ I − .
=
(A+ ×A+ )∪(A− ×A− )∪(A+ ×A− )∪(A− ×A+ )
¡ +¢ +
¡ −¢ − ¡ ¡ −¢ +
There are four possible choices of the signs ± and ±0 , yielding (where we have used m (A− )+
m (A+ ) = 1)
¡ ¢
¡ ¢
(+, +0 ) : 2m A+ I + + 2m A− I −
¡ ¡ ¢
¡ ¢¢
(+, −0 ) : 2I + + 2 m A− − m A+ I −
¡ ¡ ¢
¡ ¢¢
(−, +0 ) : 2 m A+ − m A− I + + 2I −
¡ ¢
¡ ¢
(−, −0 ) : 2m A+ I + + 2m A− I −
Note that (+, +0 ) and (−, −0 ) yield the same result. Thus,


Z 1Z 1
 2m (A+ ) I + + 2m (A− ) I − , 
|f (x) + f (y)| dxdy ≥ max 2I + + 2 (m (A− ) − m (A+ )) I − ,


0
0
2 (m (A+ ) − m (A− )) I + + 2I −
To show that
Z 1Z
0
0
Z 1Z
0
1
0
1
|f (x) + f (y)| dxdy ≥
Z
0
¡
¢
|f (x) + f (y)| dxdy − I + + I − ≥ 0,
1
|f (x)| dx, or equivalently
there are four possible cases of inequalities to consider for the pairs (I + , I − ) and (m (A+ ) , m (A− ))
:
Case 1: If I + ≥ I − and m (A+ ) ≥ m (A− ) , then
¡ ¢
¡ ¢
¡
¢
2m A+ I + + 2m A− I − − I + + I −
¡ ¡ ¢
¢
¢
¡ ¡ ¢
= 2m A+ − 1 I + + 2m A− − 1 I −
¡ ¡ ¢
¡ ¡ ¢
¢
¢
≥ 2m A+ − 1 I + + 2m A− − 1 I +
¡¡ ¡ ¡ + ¢
¡ ¢¢
¢¢
≥
2 m A + m A− − 2 I + = 0.
11
Case 2: If I + ≤ I − and m (A+ ) ≤ m (A− ) , then
¡ ¢
¡
¢
¡ ¢
2m A+ I + + 2m A− I − − I + + I −
¡ ¡ ¢
¡ ¡ ¢
¢
¢
= 2m A+ − 1 I + + 2m A− − 1 I −
¢
¢
¡ ¡ ¢
¡ ¡ ¢
≥ 2m A+ − 1 I − + 2m A− − 1 I +
¡¡ ¡ ¡ + ¢
¡ ¢¢
¢¢
≥
2 m A + m A− − 2 I − = 0.
Case 3: If I + ≥ I − and m (A+ ) ≤ m (A− ) , then m (A+ ) ≤ 12 and m (A− ) ≥ 12 , and so
¡ ¡ ¢
¡
¡ ¢¢
¢
2I + + 2 m A− − m A+ I − − I + + I −
¡ ¡ ¢
¡ ¢
¢
= I + + 2 m A− − m A+ − 12 I −
¡ ¢
≥ I + − 2m A+ I − ≥ I + − I − ≥ 0.
Case 4: If I + ≤ I − and m (A+ ) ≥ m (A− ) , then m (A+ ) ≥ 12 and m (A− ) ≤ 12 , and so
¡
¡ ¢¢
¢
¡ ¡ ¢
2 m A+ − m A− I + + 2I − − I + + I −
¡ ¢
¢
¡ ¡ ¢
= 2 m A+ − m A− − 12 I + + I −
¡ ¢
≥ −2m A− I + + I − ≥ −I + + I − ≥ 0.
12